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Current Electricity

JEE Main level, single correct, medium

354 of 380

Question 354 of 380

A moving coil galvanometer, having a resistance $G,$ produces full scale deflection when a current ${I}_{G}$ flows through it. This galvanometer can be converted into (i) an ammeter of range 0 to ${I}_{o}\left({I}_{0}\gt {I}_{g}\right)$ by connecting a shunt resistance ${R}_{A}$ to it and (ii) into a voltmeter of range 0 to $V\left(V=G{I}_{0}\right)$ by connecting a series resistance ${R}_{V}$ to it. Then,

Show answer and solution

Answer: (D) ${R}_{A}{R}_{V}={G}^{2}$ and $\dfrac{{R}_{A}}{{R}_{V}}={\left(\dfrac{{I}_{g}}{{I}_{0}-{I}_{g}}\right)}^{2}$

For ammeter:

\left({I}_{0}-{I}_{G}\right)\times {R}_{A}={I}_{G}\times G\Rightarrow {I}_{0}={I}_{G}\times \dfrac{\left(G+{R}_{A}\right)}{{R}_{A}}

For voltmeter:

{V}_{0}={I}_{G}\left(G+{R}_{V}\right)

∵{V}_{0}={I}_{0}\times G

\Rightarrow {I}_{G}\times \left(G+{R}_{V}\right)={I}_{G}\times \dfrac{\left(G+{R}_{A}\right)\times G}{{R}_{A}}

\Rightarrow {R}_{A}\times G+{R}_{V}\times {R}_{A}={G}^{2}+{R}_{A}\times G

\Rightarrow {R}_{V}{R}_{A}={G}^{2}

Again,

\dfrac{{R}_{A}}{{R}_{V}}=\dfrac{{I}_{G}\times G}{{I}_{0}-{I}_{G}}\times \dfrac{{I}_{G}}{\left({I}_{0}-{I}_{G}\right)\times G}=\dfrac{{{I}_{G}}^{2}}{{\left({I}_{0}-{I}_{G}\right)}^{2}}

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JEE Main 2025 Apr 2 Shift 1 question paper

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